# Thread: Calculus Circle Word Problem

1. ## Calculus Circle Word Problem

These word problems are driving me crazy.....

Clyde and CeCe are running around a circular track of radius 50 meters in the same direction. Clyde is running at a rate of 1.3 m/sec and CeCe is running at a rate of 0.9 m/sec. When Clyde is behind CeCe as pictured and the (straight) distance between them is 82 meters, how is the (straight) distance between them changing?

2. Let's try using the formula for a chord and the formula $s=r{\theta}$

chord length is given by $L=2rsin(\frac{\theta}{2})$

Arc length is given by $s=r{\theta}$

Clyde is catching up to CeCe, so their distance along the arc is decreasing at -0.4 m/sec.

That is ds/dt = -.4.

$\frac{ds}{dt}=r\frac{d{\theta}}{dt}+{\theta}\frac{ dr}{dt}$

$-.4=50\frac{d{\theta}}{dt}+0$. dr/dt is 0 because the radius does not change.

$\frac{d{\theta}}{dt} = -.008$

When they are 82 feet apart along the chord, then using $2rsin(\frac{\theta}{2})$

${\theta}=2sin^{-1}(\frac{82}{100})=1.92282 \;\ rad.$

So, let's differentiate implicitly our chord formula.

$\frac{dL}{dt}=50cos(\frac{\theta}{2})\frac{d{\thet a}}{dt}$

$\frac{dL}{dt}=50cos(\frac{1.92282}{2})(-.008)=-.229$

The distance between them is decreasing at .229 m/sec.

Let's see if anyone concurs. Check my work.